How can it calculated $ \lim\limits_{t \to 0}\frac{1}{t^2}\int_0^t f(g_1(x,t),g_2(x,t))\frac{\partial g_1}{\partial x_1}(x,t)-f(g_1(x,0),g_2(x,0))\frac{\partial g_1}{\partial x_1}(x,0)dx$ ,where $f,g_1,g_2:\mathbb{R}^2\to\mathbb{R}$ are $C^\infty$ function.
I guess it will be $\frac{\partial f}{\partial x_2}(0,0)\frac{\partial g_1}{\partial x_1}(0,0)\frac{\partial g_2}{\partial x_2}(0,0)$ since the formula means
"a second order terms of Integral of differential 1-form $fdx$ over a boundary of infinitesimal rectangle with sides parallel to the y-axis" and it is equal to $-\frac{\partial f}{\partial x_2}dx_1\wedge dx_2 (\frac{\partial g}{\partial x_1}, \frac{\partial g}{\partial x_2})$.
However it is difficult to solve it for me because there are $t$ inside $g_i(x,t)$.
I would appreciate for your answer.
Since $\lim\limits_{t\to 0}\frac1{t^2}{\int_0^t (F(x,t)-F(x,0))}dx=\frac{\partial F}{\partial x_2}(0,0)$, the answer is
$\frac{\partial f}{\partial x_1}(g_1(0,0),g_2(0,0))\frac{\partial g_1}{\partial x_2}(0,0)\frac{\partial g_1}{\partial x_1}(0,0)+\frac{\partial f}{\partial x_2}(g_1(0,0),g_2(0,0))\frac{\partial g_2}{\partial x_2}(0,0)\frac{\partial g_1}{\partial x_1}(0,0)+f(g_1(0,0),g_2(0,0))\frac{\partial^2 g_1}{\partial x_1 \partial x_2}(0,0)$
Furthermore, when the infinitesimal rectangle's sides are parallel to the y-axis, $\frac{\partial g_1}{\partial x_2}(0,0)=0$ then the value turs out to be $\frac{\partial f}{\partial x_2}(g_1(0,0),g_2(0,0))\frac{\partial g_2}{\partial x_2}(0,0)\frac{\partial g_1}{\partial x_1}(0,0)$ which I guess.
(A term $\frac{\partial f}{\partial x_2}(0,0)$ is actually $\frac{\partial f}{\partial x_2}(g_1(0,0),g_2(0,0))$ though)